Ag-Ag - -KI- Ag-Ag emf is E- then Ksp of AgI is given as

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Solubility Product

Ag|Ag + ||KI|...

Question

$A g | A g^{+} | | K I | A g | A g emf is E^{\circ}, then K_{s p}$ of AgI is given as

K_{s p} = \frac{n F}{2.303 R T} log E^{\circ}

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ln K_{s p} = n F [\frac{o E^{\circ}}{o T} - E^{\circ}]

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K_{s p} = \frac{n F}{E^{\circ}}

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log

K_{s p} = \frac{n F E^{\circ}}{2.303 R T}

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Similar questions

A g | A g^{+}, K I | | A g I | A g

E

K_{s p}

A g I

25^{0} C

A g + I^{-} \to A g I + e^{-}; E^{0} = 0.152 V

A g \to A g^{+} + e^{-}; E^{0} = - 0.800 V

K_{s p}

A g I

(G i v e n : 2.303 R T / F = 0.059 V)

25^{o}

A g_{(s)} + I_{(a q)} \to A g I_{(s)} + e^{-}, E^{o} = 0.152

A g_{(s)} \to A g_{(a q)}^{+} + e^{-}, E^{o} = - 0.800

K_{s p}

(W h e r e K_{s p} = solubility product, 2.303 \frac{R T}{F} = 0.059 V)

Given the data $25^{0} C$

A g + I^{-} \to A g I + e^{-}, E^{0} = 0.152 V

A g \to A g^{+} e^{-}, E^{0} = 0.800 V

What is the value of log $K_{s p}$ for AgI?

A g | A g N O_{3} (0.1 M) | K B r (1 N), A g B r (s) | A g

- 0.6 V

298 K

K_{s p}

A g B r

298 K

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